I have seen in many places that the other form of an oracle (the XOR oracle if you will) is "equivalent" to $O_f^{\pm}$
What you've seen is probably the other way: going from the other oracle to $O_f^{\pm}$. In order to do the other way around, you would have to have access to a controlled version of your oracle.
Furthermore, if you define $\overline{f}$ to be $f$ but bit-flipped:
$$\overline{f}(x)=f(x)\oplus1$$
Then $O_f^\pm$ and $O_\overline{f}^\pm$ differ only by a global phase, and thus are indistinguishable. Thus, it's not possible to recover the value of $f(x)$, as performing the same experiment with the exact opposite function would yield the same result. Note that this also tells you that you can't transform this oracle into a XOR one without having access to its controlled version.
However, what you can do is know $f(x)\oplus f(0)$. That is, if you arbitrarily fix the value of $f(0)$, then you can determine the value of $f(x)$.
Let $U$ be any unitary such that $$U|0\rangle=
\frac{|0\rangle+|x\rangle}{\sqrt2}.$$
Prepare the state $U|0\rangle$:
$$\frac{|0\rangle+|x\rangle}{\sqrt2}$$
then apply the oracle:
$$\frac{(-1)^{f(0)}|0\rangle+(-1)^{f(x)}|x\rangle}{\sqrt2}$$
and finally apply $U^\dagger$. You can convince yourself that if $f(0)=f(x)$, then you will get back $|0\rangle$ (up to a global phase). However, if $f(0)\neq f(x)$, then you'll get a state that is orthogonal to $|0\rangle$, since unitaires preserve scalar product.
Thus, if you measure $|0\rangle$, then you know that $f(x)=f(0)$. Otherwise, you know that $f(x)=f(0)\oplus1$. Of course, there's nothing special about $0$, you can fix any other arbitrary input and find whether other coefficients match its sign.
